The triangle $\triangle ABC$ is an isosceles triangle where $AB = 4\sqrt{2}$ and $\angle B$ is a right angle. If $I$ is the incenter of $\triangle ABC,$ then what is $BI$?

Express your answer in the form $a + b\sqrt{c},$ where $a,$ $b,$ and $c$ are integers, and $c$ is not divisible by any perfect square other than $1.$
Solution: We might try sketching a diagram: [asy]
pair pA, pB, pC, pI;
pA = (-1, 0);
pB = (0, 0);
pC = (0, 1);
pI = (-0.2929, 0.2929);
draw(pA--pB--pC--pA);
draw(pI--pB);
draw(circle(pI, 0.2929));
label("$A$", pA, SW);
label("$B$", pB, SE);
label("$C$", pC, NE);
label("$I$", pI, NE);
[/asy] Since $\triangle ABC$ is isosceles, we might try extending $BI$ to meet $AC$ at $D.$ That is advantageous to us since it will also be the perpendicular bisector and median to side $AC.$ In addition, let us draw a radius from $I$ that meets $AB$ at $E.$ [asy]
pair pA, pB, pC, pD, pE, pI;
pA = (-1, 0);
pB = (0, 0);
pC = (0, 1);
pD = (-0.5, 0.5);
pE = (-0.2929, 0);
pI = (-0.2929, 0.2929);
draw(pA--pB--pC--pA);
draw(pI--pB);
draw(pI--pD);
draw(pI--pE);
draw(circle(pI, 0.2929));
label("$A$", pA, SW);
label("$B$", pB, SE);
label("$C$", pC, NE);
label("$I$", pI, NE);
label("$D$", pD, NW);
label("$E$", pE, S);
[/asy] Given $r$ as the inradius, we can see that $DI = r$ and $IB = r\sqrt{2},$ since $\triangle IEB$ is also a little isosceles right triangle on its own. Therefore, $BD = r\sqrt{2} + r = r (\sqrt{2} + 1).$

However, we have a nice way of finding $BD,$ from $\triangle ABD,$ which is also an isosceles right triangle, thus $DB = \frac{AB}{\sqrt{2}} = \frac{4 \sqrt{2}}{\sqrt{2}} = 4.$

Setting the two expressions for $DB$ equal, we have: \begin{align*}
r(\sqrt{2} + 1) &= 4 \\
r &= \frac{4}{\sqrt{2} + 1} = \frac{4}{\sqrt{2} + 1} \cdot \frac{\sqrt{2} - 1}{\sqrt{2} - 1} \\
&= \frac{4(\sqrt{2} - 1)}{1} = 4\sqrt{2} - 4.
\end{align*} Our answer is $BI = r\sqrt{2} = (4\sqrt{2} - 4)\cdot \sqrt{2} = \boxed{8 - 4\sqrt{2}}.$